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Chapter 14: Problem 2092

When 100 volt dc is applied across a coil, a current of \(1 \mathrm{~A}\) flows through it. When 100 volt ac at 50 cycle \(\mathrm{s}^{-1}\) is applied to the same coil, only \(0.5\) A current flows. The impedance of the coil is, (a) \(100 \Omega\) (b) \(200 \Omega\) (c) \(300 \Omega\) (d) \(400 \Omega\)

Short Answer

Expert verified
The impedance of the coil is approximately 200 Ω, which corresponds to option (b).

Step by step solution

01

Find the Resistance (R)

We are given that a 100 V DC voltage causes a 1 A current to flow through the coil. Ohm's law states that V = IR, where V is voltage, I is current, and R is resistance. Therefore, the resistance of the coil can be found using the formula R = V/I. Plugging in the given values for voltage (100 V) and current (1 A), we calculate the resistance R: R = 100 V / 1 A = 100 Ω
02

Find the AC Current (Iac)

We are given that a 100 V, 50 Hz AC voltage causes a 0.5 A current to flow through the coil. This is the AC current (Iac): Iac = 0.5 A
03

Find the Inductive Reactance (XL)

We know the AC current and the AC voltage, so we can use Ohm's law again in the AC case to find the total impedance (Zt) of the coil: Zt = V/Iac Zt = 100 V / 0.5 A = 200 Ω The total impedance Zt is comprised of the resistance R and the inductive reactance XL and follows the equation Zt^2 = R^2 + XL^2. We can thus find the inductive reactance as follows: XL = sqrt(Zt^2 - R^2) XL = sqrt((200 Ω)^2 - (100 Ω)^2) = sqrt(30000) ≈ 173.2 Ω
04

Calculate the Impedance (Z)

Now that we have the resistance R and the inductive reactance XL, we can calculate the impedance using the formula Z = sqrt(R^2 + XL^2): Z = sqrt((100 Ω)^2 + (173.2 Ω)^2) ≈ 199.4 Ω Rounding down to the nearest whole number, we get Z ≈ 199 Ω; as it's close to one of the given options: The impedance of the coil is approximately 200 Ω, which corresponds to option (b).

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Most popular questions from this chapter

A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V}\). The current reaches half of its steady state value in \(\ldots \ldots\) (a) \(0.15 \mathrm{sec}\) (b) \(0.3 \mathrm{sec}\) (c) \(0.05 \mathrm{sec}\) (d) \(0.1 \mathrm{sec}\)

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